Question

Tìm x:
a) x(x-2)-x+2=0
b) (3-2x)²-(x-1)²=0
c) 81x⁴-x²=0
d) x³+x²+27x+27=0

Answer 1

a) \(x\left(x-2\right)-x+2=0\)

\(x\left(x-2\right)-\left(x-2\right)=0\)

\(\left(x-1\right)\left(x-2\right)=0\)

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

Answer 2

a) 

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

Answer 3

b) \(\left(3-2x\right)^2-\left(x-1\right)^2=0\)

\(\left(3-2x-x+1\right)\left(3-2x+x-1\right)=0\)

\(\left(3-3x+1\right)\left(3-x-1\right)=0\)

TH1:3-3x+1=0⇒x\(=\dfrac{4}{3}\)

TH2:3-x-1=0⇒x=2

Answer 4

a. x(x - 2) - x + 2 = 0

<=> x(x - 2) - (x - 2) = 0

<=> (x - 1)(x - 2) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

b. (3 - 2x)² - (x - 1)² = 0

<=> (3 - 2x - x + 1)(3 - 2x + x - 1) = 0

<=> (4 - 3x)(2 - x) = 0

<=> \(\left[{}\begin{matrix}4-3x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

c. 81x⁴ - x² = 0

<=> x²(81x² - 1) = 0

<=> x²\(\left[\left(9x\right)^2-1^2\right]=0\)

<=> x²(9x - 1)(9x + 1) = 0

<=> \(\left[{}\begin{matrix}x^2=0\\9x-1=0\\9x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{9}\\x=\dfrac{-1}{9}\end{matrix}\right.\)

d. x³ + x² + 27x + 27 = 0

<=> x²(x + 1) + 27(x + 1) = 0

<=> (x² + 27)(x + 1) = 0

<=> \(\left[{}\begin{matrix}x^2+27=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{27}\\x=-1\end{matrix}\right.\)